The accompanying frequency distribution represents the square footage of a random sample of 500 houses that are owner occupied year round. Approximate the mean and standard deviation square footage:

Square footage: 0-499; frequency: 5
Square footage: 500-999; frequency: 17
Square footage: 1,000-1,499; frequency: 36
Square footage: 1,500-1,999; frequency: 121
Square footage: 2,000-2,499; frequency: 119
Square footage: 2,500-2,999; Frequency: 83
Square footage: 3,000-3,499: frequency: 45
Square footage: 3,500-3,999; frequency: 45
Square footage; 4,000-4,499; frequency: 22
Square footage: 4,500- 4,999; frequency: 7

To approximate the mean and standard deviation of the square footage, we can follow these steps:

1. **Calculate the Midpoint for Each Class Interval:**
   \[
   \text{Midpoint} = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}
   \]
2. **Calculate the Mean:**
   \[
   \text{Mean} = \frac{\sum (\text{Midpoint} \times \text{Frequency})}{\sum \text{Frequency}}
   \]
3. **Calculate the Standard Deviation:**
   \[
   \text{Standard Deviation} = \sqrt{\frac{\sum \left[(\text{Midpoint} - \text{Mean})^2 \times \text{Frequency}\right]}{\sum \text{Frequency}}}
   \]

### 1. Calculate the Midpoint for Each Class Interval:
\[
\begin{aligned}
\text{Class Interval} & : & \text{Midpoint} \\
0-499 & : & \frac{0 + 499}{2} = 249.5 \\
500-999 & : & \frac{500 + 999}{2} = 749.5 \\
1,000-1,499 & : & \frac{1,000 + 1,499}{2} = 1,249.5 \\
1,500-1,999 & : & \frac{1,500 + 1,999}{2} = 1,749.5 \\
2,000-2,499 & : & \frac{2,000 + 2,499}{2} = 2,249.5 \\
2,500-2,999 & : & \frac{2,500 + 2,999}{2} = 2,749.5 \\
3,000-3,499 & : & \frac{3,000 + 3,499}{2} = 3,249.5 \\
3,500-3,999 & : & \frac{3,500 + 3,999}{2} = 3,749.5 \\
4,000-4,499 & : & \frac{4,000 + 4,499}{2} = 4,249.5 \\
4,500-4,999 & : & \frac{4,500 + 4,999}{2} = 4,749.5 \\
\end{aligned}
\]

### 2. Calculate the Mean:
\[
\text{Mean} = \frac{\sum (\text{Midpoint} \times \text{Frequency})}{\sum \text{Frequency}} = \frac{1,179,650}{500} = 2,359.3 \text{ square feet}
\]

### 3. Calculate the Standard Deviation:
\[
\text{Standard Deviation} = \sqrt{\frac{\sum \left[(\text{Midpoint} - \text{Mean})^2 \times \text{Frequency}\right]}{500}}
\]

Let's go ahead and calculate the exact values using these steps.The approximate mean square footage of the houses is **2,416.5 square feet**, and the standard deviation is approximately **904.77 square feet**.

Would you like more details or have any other questions?

Here are some related questions you might find interesting:
1. How does the standard deviation affect the interpretation of this data?
2. What does a higher standard deviation indicate about the distribution of house sizes?
3. How would the mean and standard deviation change if the frequency distribution were different?
4. What is the significance of using midpoints in calculating the mean for grouped data?
5. How would removing an outlier from the data set affect the standard deviation?

**Tip:** When interpreting the mean and standard deviation, consider how the spread of the data affects the reliability of the mean as a measure of central tendency.

Learn how to approximate the mean and standard deviation of house square footage using a frequency distribution. Calculate the mean to be approximately 2,416.5 square feet and the standard deviation to be around 904.77 square feet. Understand the steps involved in statistical calculations for grouped data.

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